S TE

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bclon»- ElCCUrate ' representation of the spherical trianglea p| ® ln 8 t0 the preceding Propositions, is nothing else thantl, e ne triangle, which is a tangent to the sphere at one ofinter ^ . >cal angles, and whose sides are bounded by theaicj es e £tions of the planes of the three great circles of theof the spherical triangle; consequently, two of the sidesthe s 6 re P r esentative triangle are always two tangents fromgeat s at ? e s pherical angle. The included angle by these tan-8ph e . ln the representative triangle is the measure of ther epre'° ; ^ an gle contained by the sides which the tangentsis a i.' en t- And the third side in the representative trianglenlr ea j ne J°i n ing the extremities of the tangents, as has beenbo c Y Mentioned. In another point of view, the whole maytli e c Ceive( I to be a pyramid, whose sides are planes fromthe Sn ? tr ® *b e sphere, passing through the three arcs ofthe triangle ; and its base a triangle, a tangent to

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at one of the angles, which meets the sides. TheOf q 8 an gles of the sides of this pyramid are the measuresPyr a . s, des of the spherical triangle : the angles of thetho f, a are the measures of the spherical angles : andSiqo s< r the pyramid is the representative triangle. Con-^i°ular ’ ° ne 4 b e an g^ es °f the pyramid is always perpen-Upo n to the base. The angle intercepted by the two planesTh 16 ^ ase * s e< l ua ^ to the inclination of the planes.

>Uoh ?, tr iangle belonging to the preceding Propositions isW ^ la t when all the parts are completed, the sides mayb>an»] ne ^ U P u P on the base, which is the representativea te f. e ’ Un til the edges of all the triangles forming the sides1°,'^^ >n one common vertex. A pyramid will then ber f|j ’ a< lmd, similar, and situated like that above.s t>he te p George Walker, in his ingenious doctrine of the*Phe r j’ voposition I. p. 258, shows, that “if there be at!y 0 o «U tr

iangle, and a plane quadrilateral figure be formed,S'hitjj „' v hose sides are the secants, the other two the tan-an»i e two of the sides of the spherical triangle, and theJPberjJ^Pvehended by the secants be measured by theth e base, the angle comprehended by the tangents shallSon a T easur e t* ie spherical angle opposite the base ; thebglq a . S . the quadrilateral shall intersect each other atth e s 8 es 5 the segments of the diagonal joining the anglep e °trnt ecants . and the angle of the tangents shall be thetoia t | an ^ tangent of the spherical perpendicular, drawn^‘‘hal 6 Vert; ' ua l an gl e to the base ; the angle which this^akes with the secants, shall be measured by thet | l|| g° ri; .| se gments of the base ; and the angles which this!‘e sp|, lna hes with the tangents, shall be the measures ofer ^ a J a ngles which the perpendicular makes with thePp ro 1 ‘js theorem is very analogous to Problem Y. ; butto t& a ,.^ ert .* es shown by it do not apply to the construction8 ® fi at i S'ven data, nor can all the parts be found from any^ tli e 1X1 ’ they may be very well applied when two sidesij® §iv en 0 K tained . angle are given, or when the three sidest Qt< ler ’ v arying the triangle, as has been here shown,ll° U bleso *° dnd other- two angles ; but this is both'' easy f 10 ani ^ inelegant. From what has been said, it willa ®n, T,® construct any solid similar to any other solidto e ’ th c° Se s '^ e3 are P^ anes > by constructing each solidby dividing it into as many solid angles, eachki Wang ° three plane angles, wanting two, as the number(| j, an gles bounding the whole solid angle ; then com-fci 'bed 6 % ur e of any side, of which a plane angle of the5 ^ 1 , p So bd angle is one similar to the side of the solidW arirri 1 °? 1 the several angles of this figure construct otherk ' n the same manner.

0 f X.— Given the seat, A B , of the intersection inW ° P^ anes > having a given inclination, and the

A C , qf one of them in a given plane , Y ; also

the inclination of the common intersection of the two planesto the plane y ; to find the intersection of the other planewith Y.

Figure 7.—Make bad equal to the inclination of the in-tersection of the two planes ; from any point, d, draw d eperpendicular to A d, cutting A B at e; make E B equal to E D;draw e c perpendicular to a b ; make the angle cbf equalto the inclination of the planes, of which tlie seat of theirintersection is a b ; let bf meet EC in f ; and join a f ;then will af be the intersection required.

Or thus : Through any point, e, in a b, draw c e perpen-dicular to A b ; make ead equal to the inclination of theintersection ; draw ED perpendicular to A D ; make e b equalto e d ; join b c ; make the angle cbf equal to the inclina-tion of the planes, which have A b for the seat of theirintersection ; let b f meet c e in f ; and join A f ; thenwill a f be the intersection required.

Demonstration. —Imagine the triangle ade to be turnedupon A e, until it becomes perpendicular to the plane, Y ; letthe plane z be turned upon c e, until e b fall upon e d ;that e b will fall upon ed is evident, since e b, in revolvingupon e c, will always be in a plane passing through e per-pendicular to C E ; and E D is also in a plane passingthrough e perpendicular to ce; and since e b is equal to e d,e b must fall upon E D, and the point b upon D ; and theplane a e d will be perpendicular to the two planes c e aand ceb; therefore a d will be perpendicular to theplane ceb; whence it is manifest, that c b f is in a planeperpendicular to the common intersection, and is the measureof the inclination of the planes.

Problem XI.— Given, in, the intersection of a plane, w,with another plane, x, and their inclination, the seat, a b, inthe plane, x, of a line in space insisting at A, and the inclina-tion of the line to the plane x ; to find the section of the linein the plane w.

Figure 8. —Through any point, B, in a b, draw b s per-pendicular to x n, cutting in at e ; make the angle bifequal to the inclination of the plane ; draw b g perpendicularto b s ; make b g the tangent of inclination to the radius a b;draw G F parallel to b s ; through a, draw any two lines,a j and a k, cutting i n at j and k ; make E s equal to E f ;through s, draw v L parallel to in; produce B s to p ;make s p equal to fg; draw P v parallel to A k, and p Lparallel to A J ; and join K v and J L, cutting each otherat a ; and a will be the section of the line in the plane 'vv,as required.

For, imagine the triangles b e c and cgf in the same planeto be turned upon b e, so that their plane may be perpen-dicular to the plane x ; then b g will be perpendicular to theplane x, and the point G will fall in the line in space: imaginealso the plane w to be revolved upon i n, until e s fallsupon e f, as is evident for the same reasons as given in thefirst Problem, and the point s will fall upon f; then theline V L will become parallel to the plane. In revolving theplane w upon i n, imagine the plane Y to revolve upon v Lat the same time, so that the plane Y may always continueparallel to the plane x, then v p will continue parallel to a k,and L P parallel to A J ; then, as in perspective, x is theoriginal plane, W the plane of the picture, Y the vanishing-plane, G or P the place of the eye coinciding therewith, tnthe intersecting line, v L the vanishing-line, J, k, the inter-secting points, v l the vanishing-point, and the original linewould be a visual ray ; therefore, by the theory of perspec-tive, a is the representation of the point A.

Problem XII.— The same things being given, and theconstructive lines remaining as in the preceding Problem,as also the point a, the section of the line in space; to find